CJC For Sharing_2012 JC2 H2 Prelim Paper 1 Solutions

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 1 INEQUALITIES + COMPLEX NUMBERS Part Assessment Objectives Mark Scheme Feedback (i) - solve an equality involving modulus functions either algebraically or graphically - deduce the solution of an inequality of the form 0)( )( xg xf where f(x) and g(x) are linear expressions, based on the solution obtained from solving the equality Suppose ,w       20 023 063 12144 112 112 2 22 22       w ww ww wwww ww ww   1,2,0 023 112 11 12      www ww ww w w (ii) - express a complex number in its algebraic form - represent complex numbers expressed in cartesian form by points in the Argand diagram Suppose ,Cw let iyxw  , where x, y . Since 112  ww ,   . 02 , 0363 , 124144 , || that note , )1(412 , )1(2)12( , )1(2)12( , 11)(2 2222 2222 2222222 22       yxxyxx yxxyxx babiayxyx iyxyix iyxyix iyxiyx

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 1 INEQUALITIES + COMPLEX NUMBERS Part Assessment Objectives Mark Scheme Feedback Then, via completing the square,   11 22  yx , which represents a circle of radius 1 and centred about (1, 0), together with its interior.

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 2 AP & GP Part Assessment Objectives Mark Scheme Feedback (a) Let n be the no. of months. Then 2000)10)1(100(2 2000))1(2(2   nn dnan 2000)1090(2  nn 040092  nn Solving, 16,25  nn . 16 complete months to fully repay the debt. (b)(i) Find the general of a GP and use it to find a particular value Balance at end of 1 month Balance at end of 2 months [ ] Balance at end of n months ( ) Outstanding amount at 1 January 2013 (ii) Solve inequality using general term. loan repaid ⇒ ⇒ ⇒ so he takes 78 months.

CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 9740 JC2 PRELIM PAPER 1 SOLUTIONS 2012 3 SIGMA NOTATION/METHOD OF DIFF/MI Part Assessment Objectives Mark Scheme Feedback (i) Able to prove a statement using the method of mathematical induction involving a recurrence relation. Let nP be the statement   2 1 1   n nun for all 1n . When 1n , LHS 01  u RHS   0 11 11 2    Since LHS = RHS, 1P is true and forms the basis for induction. Assume kP to be true, i.e.   2 1 1   k kuk . RTP:   21 2  k kuk . LHS     2 2 1 21 4    kk kkuu kk       2 2 2 21 4 1 1     kk kk k k                      22 2 22 23 22 223 22 22 21 12 21 2 21 443 21 421         kk kkk kk kkk kk kkkk kk kkkk   RHS